# How do you integrate 1/((x+6)(x^2+3)) using partial fractions?

Nov 5, 2016

$\frac{1}{78} \ln \left({\left(x + 6\right)}^{2} / \left({x}^{2} + 3\right)\right) + \frac{2}{13 \sqrt{3}} \arctan \left(\frac{x}{\sqrt{3}}\right) + C$

#### Explanation:

Decomposing the fraction:

$\frac{1}{\left(x + 6\right) \left({x}^{2} + 3\right)} = \frac{A}{x + 6} + \frac{B x + C}{{x}^{2} + 3}$

So:

$1 = A \left({x}^{2} + 3\right) + \left(B x + C\right) \left(x + 6\right)$

$1 = A {x}^{2} + 3 A + B {x}^{2} + 6 B x + C x + 6 C$

Sorting by $x$:

$0 {x}^{2} + 0 x + 1 = {x}^{2} \left(A + B\right) + x \left(6 B + C\right) + \left(3 A + 6 C\right)$

Comparing the coefficients:

$\left\{\begin{matrix}A + B = 0 \\ 6 B + C = 0 \\ 3 A + 6 C = 1\end{matrix}\right.$

From the first equation we see that $B = - A$. Substituting this into the second equation, we see that $- 6 A + C = 0$. Multiplying the third equation by $2$, we see that $6 A + 12 C = 2$. Adding this to $- 6 A + C = 0$, we get $13 C = 2$ so $C = \frac{2}{13}$.

Using this value in the second equation, we see that $A = \frac{1}{39}$. Thus $B = - \frac{1}{39}$.

Using these values:

$\frac{1}{\left(x + 6\right) \left({x}^{2} + 3\right)} = \frac{\frac{1}{39}}{x + 6} + \frac{- \frac{1}{39} x + \frac{2}{13}}{{x}^{2} + 3}$

$\frac{1}{\left(x + 6\right) \left({x}^{2} + 3\right)} = \frac{1}{39} \frac{1}{x + 6} + \frac{1}{39} \frac{- x + 6}{{x}^{2} + 3}$

So:

$I = \int \frac{1}{\left(x + 6\right) \left({x}^{2} + 3\right)} \mathrm{dx} = \frac{1}{39} \int \frac{1}{x + 6} \mathrm{dx} + \frac{1}{39} \int \frac{- x + 6}{{x}^{2} + 3}$

And:

$I = \frac{1}{39} \int \frac{1}{x + 6} \mathrm{dx} - \frac{1}{39} \int \frac{x}{{x}^{2} + 3} \mathrm{dx} + \frac{6}{39} \int \frac{1}{{x}^{2} + 3} \mathrm{dx}$

Letting $u = x + 6$ for the first integral, so that $\mathrm{du} = \mathrm{dx}$:

$I = \frac{1}{39} \int \frac{1}{u} \mathrm{du} - \frac{1}{39} \int \frac{x}{{x}^{2} + 3} \mathrm{dx} + \frac{2}{13} \int \frac{1}{{x}^{2} + 3} \mathrm{dx}$

$I = \frac{1}{39} \ln \left\mid u \right\mid - \frac{1}{39} \int \frac{x}{{x}^{2} + 3} \mathrm{dx} + \frac{2}{13} \int \frac{1}{{x}^{2} + 3} \mathrm{dx}$

For the next integral, let $v = {x}^{2} + 3$ so $\mathrm{du} = 2 x \mathrm{dx}$:

$I = \frac{1}{39} \ln \left\mid u \right\mid - \frac{1}{78} \int \frac{2 x}{{x}^{2} + 3} \mathrm{dx} + \frac{2}{13} \int \frac{1}{{x}^{2} + 3} \mathrm{dx}$

$I = \frac{1}{39} \ln \left\mid u \right\mid - \frac{1}{78} \int \frac{1}{v} \mathrm{dv} + \frac{2}{13} \int \frac{1}{{x}^{2} + 3} \mathrm{dx}$

$I = \frac{1}{39} \ln \left\mid u \right\mid - \frac{1}{78} \ln \left\mid v \right\mid + \frac{2}{13} \int \frac{1}{{x}^{2} + 3} \mathrm{dx}$

Before moving onto the next integral, we can do a very sneaky simplification:

$I = \frac{2}{78} \ln \left\mid u \right\mid - \frac{1}{78} \ln \left\mid v \right\mid + \frac{2}{13} \int \frac{1}{{x}^{2} + 3} \mathrm{dx}$

$I = \frac{1}{78} \ln \left\mid {u}^{2} \right\mid - \frac{1}{78} \ln \left\mid v \right\mid + \frac{2}{13} \int \frac{1}{{x}^{2} + 3} \mathrm{dx}$

$I = \frac{1}{78} \ln \left\mid {u}^{2} / v \right\mid + \frac{2}{13} \int \frac{1}{{x}^{2} + 3} \mathrm{dx}$

With $u = x + 6$ and $v = {x}^{2} + 3$:

$I = \frac{1}{78} \ln \left\mid {\left(x + 6\right)}^{2} / \left({x}^{2} + 3\right) \right\mid + \frac{2}{13} \int \frac{1}{{x}^{2} + 3} \mathrm{dx}$

Note the absolute value bars aren't necessary since the function inside the logarithm is always positive:

$I = \frac{1}{78} \ln \left({\left(x + 6\right)}^{2} / \left({x}^{2} + 3\right)\right) + \frac{2}{13} \int \frac{1}{{x}^{2} + 3} \mathrm{dx}$

For the remaining integral, there are two courses of action. The first would be to use the arctangent integral formula: $\int \frac{1}{{u}^{2} + {a}^{2}} \mathrm{du} = \frac{1}{a} \arctan \left(\frac{u}{a}\right) + C$.

The other, which I prefer since you don't have to remember the formula, is to use trigonometric substitution.

So, for $\int \frac{1}{{x}^{2} + 3} \mathrm{dx}$, let $x = \sqrt{3} \tan \theta$. Thus $\mathrm{dx} = \sqrt{3} {\sec}^{2} \theta d \theta$ and:

$I = \frac{1}{78} \ln \left({\left(x + 6\right)}^{2} / \left({x}^{2} + 3\right)\right) + \frac{2}{13} \int \frac{1}{3 {\tan}^{2} \theta + 3} \left(\sqrt{3} {\sec}^{2} \theta d \theta\right)$

$I = \frac{1}{78} \ln \left({\left(x + 6\right)}^{2} / \left({x}^{2} + 3\right)\right) + \frac{2}{13 \sqrt{3}} \int {\sec}^{2} \frac{\theta}{{\tan}^{2} \theta + 1} d \theta$

Since ${\sec}^{2} \theta = 1 + {\tan}^{2} \theta$:

$I = \frac{1}{78} \ln \left({\left(x + 6\right)}^{2} / \left({x}^{2} + 3\right)\right) + \frac{2}{13 \sqrt{3}} \int d \theta$

From $x = \sqrt{3} \tan \theta$ we see $\theta = \arctan \left(\frac{x}{\sqrt{3}}\right)$.

$I = \frac{1}{78} \ln \left({\left(x + 6\right)}^{2} / \left({x}^{2} + 3\right)\right) + \frac{2}{13 \sqrt{3}} \arctan \left(\frac{x}{\sqrt{3}}\right) + C$